Two-phase squarewave oscillator


Posted on Feb 5, 2014

It`s one of the simplest circuits to get working but one of the more tricky to understand. This simple two-transistor oscillator has a technical name of astable multivibrator  and provides a squarewave (of sorts) output in and out of phase. First up, keep in your mind that capacitors cannot change their voltage instantly. So we`ll assume that th


Two-phase squarewave oscillator
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e cap attached to R2 has 0V across it. This pulls the base of Q2 to 0V, which means it`s off. However, with Q1 on, that 0. 01uF capacitor now begins to charge via R2. When the voltage at the junction of the cap and R2 (which is also the base of Q2) reaches 0. 6V. This now turns Q2 on. Because Q1 was on, that means the cap joining R3 had 0. 6V across it (it can`t go any higher because the Vbe drop of Q1 won`t let it go higher than 0. 6V) but Q2 switching on effectively pulls the side of the capacitor connected to R4 from 9V down to 0V. As I said before, capacitors cannot instantly change the voltage across them so what happens on the R3 side of capacitor C2 is its voltage goes from 0. 6V down to -8. 4V, which well and truly turns Q1 off. At the same time, a similar thing has happened to cap C1, connected to R1 and R2. The R1 side of that cap has jumped from 0V to 5VDC and the other side jumps from 0. 6V to 5. 6V well, it would if it wasn`t for the Vbe junction of Q2, which holds it to 0. 6V. So now that Q2 is on and Q1 is off, C2 now begins charging up via R3 and the Q2 collector-emitter path. But it`s starting from -8. 4V and it has to reach 0. 6V before it turns Q1 back on again. However, it will eventually do so, turning Q1 on, which pulls the R1 side of C1 from 9V down to 0V and the R2 side from 0. 6V to -8. 4V. How the circuit actually starts in practice is going to be a bit of a lottery, even with component pairs the...




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